Mathematical Induction where the base case starts above 1

1. The problem statement, all variables and given/known data
Find all natural numbers such that 2n ≥ (1+n)2, and prove your answer.

2. The attempt at a solution
I can see this is true for n=0 and n>5. I try to prove this using induction as follows
20 =1≥ 1=(1+0)2
base case: 26 =64≥ 49=(1+6)2 so it is true for n=6
and suppose 2n ≥ (1+n)2 for all n≥6 then
2n+2 =2n22 ≥4(1+n)2=4n2+8n+4≥n2+6n+9=(n+3)2

I'm not sure if this correct because of the +2?
Would I have do to it again with a base case of 7 to ensure every number is accounted for?

Clearly using n+2 was a mistake but I originally used it to see if this would result in an easier inequality for me to manipulate. I see your proof and it makes sense. I didnt see it originally but my original proof works for n+1 aswell.

2n+1 =2n2 ≥2(1+n)2=2n2+4n+2≥n2+4n+4=(n+2)2

But suppose I could not show this was true for n+1 does the logic for induction work if we use n+2. ie would the original proof using n+2 be sufficient although not elegant?